`2x^(2) + 3x - alpha = 0 " has roots "-2 and beta " while the equation "x^(2) - 3mx + 2m^(2) = 0 " has both roots positive, where " alpha gt 0 and beta gt 0.`
`Delta PQR, angleR=pi/2. " If than "(P/2) and tan (Q/2) " are the roots of the equation " ax^(2) + bx + c = 0 ,` then which one of the following is correct ?

Options

A. `a+b=c`
B. `b+c=a`
C. `c+a=b`
D. `b=c`

Answer

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Understand the given equations and roots

We have two equations:
1. \( 2x^2 + 3x - \alpha = 0 \) with roots \( -2 \) and \( \beta \).
2. \( x^2 - 3mx + 2m^2 = 0 \) with both roots positive.

### Step 2: Use Vieta's formulas for the first equation

According to Vieta's formulas, for the first equation:
- The sum of the roots \( (-2 + \beta) = -\frac{b}{a} = -\frac{3}{2} \)
- The product of the roots \( (-2 \cdot \beta) = \frac{c}{a} = -\frac{\alpha}{2} \)

From the sum of the roots:
\[ -2 + \beta = -\frac{3}{2} \implies \beta = -\frac{3}{2} + 2 = \frac{1}{2} \]

From the product of the roots:
\[ -2 \cdot \frac{1}{2} = -\frac{\alpha}{2} \implies -1 = -\frac{\alpha}{2} \implies \alpha = 2 \]

### Step 3: Analyze the second equation

Now, we look at the second equation \( x^2 - 3mx + 2m^2 = 0 \). For this equation, we need both roots to be positive.

Using Vieta's formulas again:
- The sum of the roots \( r_1 + r_2 = 3m \) (must be positive)
- The product of the roots \( r_1 \cdot r_2 = 2m^2 \) (must be positive)

Since both roots are positive, we conclude that:
- \( 3m > 0 \implies m > 0 \)
- \( 2m^2 > 0 \implies m \neq 0 \)

### Step 4: Analyze the triangle PQR

Given that triangle PQR has \( \angle R = \frac{\pi}{2} \), we can find the angles \( P \) and \( Q \):
\[ P + Q = \frac{\pi}{2} \]

### Step 5: Find the roots of the quadratic equation

The roots of the equation \( ax^2 + bx + c = 0 \) are given as \( \tan\left(\frac{P}{2}\right) \) and \( \tan\left(\frac{Q}{2}\right) \).

Using the sum and product of the roots:
- Sum of roots:
\[ \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = -\frac{b}{a} \]
- Product of roots:
\[ \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right) = \frac{c}{a} \]

### Step 6: Apply the tangent addition formula

Using the tangent addition formula:
\[ \tan\left(\frac{P}{2} + \frac{Q}{2}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus,
\[ \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = 1 \cdot \left(1 - \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right)\right) \]

### Step 7: Set up the equation

From the above, we derive:
\[ -\frac{b}{a} = 1 \quad \text{and} \quad \frac{c}{a} = \tan\left(\frac{P}{2}\right) \tan\left(\frac{Q}{2}\right) \]

### Step 8: Conclusion

After substituting and simplifying, we find that:
\[ a + b = c \]

### Final Answer

Thus, the correct relation is:
\[ \boxed{a + b = c} \]